1.101 Structures Lab.
Fall 2005
Week 2 - Beam Bending
Our objective is to compare engineering beam theory with experiment.
We will subject a “simply supported” beam to “pure bending” (over a portion of its span) and mea­
sure the strain (stress) at the top and bottom surfaces and the mid-span, vertical displacement.
Once again strain gages will be used to measure the former and a dial gage indicator employed to
measure the latter. We will restrict our loading to the elastic range.
You will use the pages that follow to report your results.
The test set-up and procedure.
We use the same specimen used in last week’s tension test only now we apply the load trans­
versely. You will discover that it takes very little load P, on the order of a pound or two to pro­
duce a significant and visible, vertical displacement. We will also test a second specimen of
double the thickness.
L
a
a
L/2
c
h
z
x
c
y
b
CL
P
The span L should be no more than 10 inches. The dial gage is to measure the displacement at
mid-span. The strain gages, on the other hand can be located off the center line as shown for the
region -c < x < +c will experience a constant bending moment. (See Appendix A for the proof of this
statement). Hence the strains at any point within that region at the top and bottom of the beam will
not change with position x.
Locate the dial gage at mid-span, making sure it is supported as rigidly as possible. You will
note that the dial will turn “backwards” as the beam deflects downward. Make sure that the dial
gage will not run out of play as the beam deflects.
The load is applied using the light-weight yoke suspended (by kevlar string) from two points located
symmetrical at x = +/- c. Make sure that the strings are located symmetrically with respect to
the center line. The distance a should be as small as feasible but you are not to interfere with the
strain gages. You will again use a gaged specimen for temperature compensation. This should rest
on the roller supports, parallel and next to the loaded specimen. (Not shown in the figure).
In making a comparison with engineering beam theory, you will need to compute the value of the
geometric parameter, I, the “second moment of area” of the cross section-often called the moment
of inertia of the cross section. (See Appendix A). Measure the width, b, and thickness, h.
b = ____________.
h = ________________
I = bh3/12 = _________________ (units)
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Vsupply (e.g., +6 volts)
R+∆R
R-∆R
e1
R
e2
R
The placement of the gages in the Wheatstone bridge is dif­
ferent from their placement in the tension test. There, in last
week’s test, the two gages were both in tension. Now, with the
loading shown, the gage at the bottom of the beam will be in ten­
sion, the top gage in compression. The gages should be located
as shown in the figure at the left in order to maximize the differ­
ential voltage, e1-e2. See appendix B for analysis of this circuit.
Follow the same procedure you used last week to determine the
gain of the op-amp, replacing one of the gages with the 2kohm
trim pot in parallel with a 360 ohm resistor.
Op-amp Gain
input = e1-e2
output
mv
volts
[+/-??]=
[+/-??]=
op-amp output voltage
input mv
Gain: G = ________
When you are finished estimating the gain, disconnect the 2 kilo-ohm trim pot and replace the
strain gage you had removed from the bridge at the outset.
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The bending test
With all four gages now connected in the bridge circuit, adjust the 10 kohm pot to bring the output to
zero, within a few millivolts or tenths of millivolts. Make sure you have positioned the dial gage indi­
cator prior to this step since the dial gage will tend to load down the beam.
Attach the light weight yoke to the strings hanging down, through the table. You will add the small
blue or better yet, the brown weights, to increment the load. Do not overload: You need to estimate
how much weight you can add and yet stay below 20% of the yield stress. (See Appendix A)
You are to take two sets of readings with each of the two specimens. For each specimen, set the
supports at two different span lengths, e.g, 10 inches, and 8 inches.
Use the set of forms below to take data and present your numerical results. (Or append a spread
sheet).
TEST 1. Beam Bending Data
h = 1/16 in; L = ____________; c = _________.
Load
op-amp output
Dial gage ind.
Displacement
Stress
[units]
[units]
[units]
[units]
[units]
+/­
+/­
+/­
+/­
+/­
Column 1
Column 2
Column 3
Column 4
Column 5
Columns 1 - 3 are measured data. Column 1 is the load; Column 2 is the op-amp output; Column 3
is the dial gage indicator reading.
Column 4, is the vertical displacement at the at the dial gage (positive downwards). Columns 5, is
the stress at the top (compression) and at the bottom (tension) of the beam, computed from the
product of the strain and the elastic modulus, E. The strain, is obtained from the sequence of rela­
tionships that relate the strain to the fractional change in resistance to the output of the bridge to the
output of the op-amp as in last week’s tension test.
Make two plots: One showing how the load (ordinate axis) varies with the displacement (abscissa);
a second, showing how the stress varies with the load (abscissa). [Generally, this is the way the
results are plotted]. Show the results of engineering beam theory on these same plots.
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TEST 1. h = 1/16 L =
c=
Load P
Mid-span Displacement
Stress σ
Load P
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Make a second test, same specimen, but move the supporting rollers inward to reduce the length of
span L. keeping the distance c the same as in the first test.
TEST 2. Beam Bending Data
thickness h = 1/16 in; L = ____________; c = _________.
Load
op-amp output
Dial gage ind.
Displacement
Stress
[units]
[units]
[units]
[units]
[units]
+/­
+/­
+/­
+/­
+/­
Column 1
Column 2
Column 3
Column 4
Column 5
Again plot the results and compare with engineering beam theory.
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TEST 2. h = 1/16 L =
c=
Load P
Mid-span Displacement
Stress σ
Load P
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Now subject the thicker specimen to bending in the same way, starting with a span length of
something less than 10 inches.
TEST 3. Beam Bending Data
thickness h = 1/8 in; L = ____________; c = _________.
Load
op-amp output
Dial gage ind.
Displacement
Stress
[units]
[units]
[units]
[units]
[units]
+/­
+/­
+/­
+/­
+/­
Column 1
Column 2
Column 3
Column 4
Column 5
Again plot the results and compare with engineering beam theory. For this, you will need to com­
pute the second moment of area of this thicker specimen;
b = ____________.
h = ________________
I = bh3/12 = _________________ (units)
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Test 3. h = 1/8 in; L=______________ c =
Load P
Mid-span Displacement
Stress σ
Load P
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For a final test, reduce the span length.
TEST 4. Beam Bending Data
thickness h = 1/8 in; L = ____________; c = _________.
Load
op-amp output
Dial gage ind.
Displacement
Stress
[units]
[units]
[units]
[units]
[units]
+/­
+/­
+/­
+/­
+/­
Column 1
Column 2
Column 3
Column 4
Column 5
Again plot the results and compare with engineering beam theory.
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Test 4. h = 1/8 in; L=______________ c =
Load P
Mid-span Displacement
Stress σ
Load P
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Appendix A
Engineering Beam Theory
For a beam loaded symmetrically with respect to
mid-span as shown, the applied loads engender a
“bending moment” distribution, My(x) and a “shear
force” distribution, Qz(x), over the length of the
beam. The convention for positive force and
moment, in accord with 1.050, is shown in the fig­
ure. Not shown are the two vertical reactions at the
points of simply-support. Each is equal in magni­
tude to F.
L
z,
F
a
c
F
c
a
x
Qz
F
Note that the shear force and bending moment dis­
tributions satisfy the equilibrium requirements:
-F
dQ z
= 0
dx
z
dM y
– Qz = 0
dx
Qz
My
My
x
In writing the first of these, which guarantees force
equilibrium in the vertical direction, we have
neglected the (uniformly distributed) weight of the
beam itself.
My=- a*F
Note that over the mid-section of the beam, the
shear force is zero and the bending moment constant. We say this section is in “pure bending”. The
bending moment within the mid-section -c< x <+c, is the product of the distance a and the force F
(as shown in class).
Engineering beam theory shows that the most
significant stress is the normal stress compo­
nent on an “x face”; σxx in the example at the
right. It is related to the bending moment by
σ xx
∫z
2
z
Qz
My
x
M ⋅z
= ------y------­I
F
In this equation, z is the distance from the
“neutral axis” which, for a doubly symmetric
beam, is at the center of the cross-section and
I is the “moment of inertia” of the cross section
I =
σxx
L
z,
F
a
a
x
dA
A
For a rectangular cross-section of width b and height h, this is I = bh3/12
h
b
The shear force engenders a shear stress, σzx which distributes over the face, directed in the verti­
cal direction (not shown) but this is of a smaller order of magnitude. We will also neglect its effect on
the displacement of the beam.
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The extensional strain component, εxx, under the same set of assumptions, is simply proportional to
the normal stress component,
σxx..
σ xx M y ⋅ z
- = ------------­ε xx = ------E
EI
For our beam, with its doubly symmetric cross-section, the (compressive) strain at the top of the
beam is equal in magnitude to the (tensile) strain at the bottom of the beam. This justifies our place­
ment of the strain gages in the Wheatstone bridge.
From the geometry of deformation, the exten­
sional strain is given by
F
ε xx = – z ⁄ R
L
z,
a
w(x)
F
a
x
where R is the radius of curvature of the axis. (1/
R) is the curvature. The negative sign is neces­
say since we take the curvature, 1/R, to be posi­
tive when the beam bends concave upwards.
From the last two relationships we see that the
bending moment is related to the radius of curva­
ture at any point along the span by
1
– M y = M b = ( EI ) ⋅  --­
R
R
Mb.
Mb= - My
(Please excuse the introduction of Mb).
The curvature, for small deflections, is related
to the vertical displacement of the axis of the
beam by, (1/R) = d2w/dx2.
Finally, an integration of the differential equation obtained from the moment curvature relation gives,
taking account of the boundary conditions that the displacement at the supports must vanish, for
the mid-span deflection
w
midspan
2
2
Fa
= –  ------------ ⋅ (3 L – 4 a )
 24 EI
NOTE: F is one half of the load added to the chain below the table surface.
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Appendix B. - The bridge circuit.
The “Wheatstone bridge” produces an output voltage proportional to the change in resistance of the
active strain gages. This voltage signal, in turn, is input into an operational amplifier, which boosts
the (dc) signal by a factor, the Gain factor. We analyze the bridge circuit anew since the placement
of the gages in the bridge circuit differs from last week.
The two strain gages fixed to the specimen subject to loading are
positioned as shown. The figure indicates that the active gage
located top left is the gage fastened to the top surface of the
beam. It is subject to compression, hence its resistance
decreases. The active gage located top right in the figure is fastened to the bottom surface of the beam and experiences tension.
Note: It does not matter if you switch the location of the two
active gages in the bridge circuit. This will just change the
polarity of the voltage difference between e1 and e2.
As before, we use two “in-active” gages for temperature compensation. We apply the circuit laws to determine the output voltage
e1 - e2 as a function of ∆R/R.
Vsupply (e.g., +6 volts)
R-∆R
R+∆R
i
e1
e2
i
R
R
Consider the current flow through the two resistors on the left side
of the bridge. From the usual circuit laws, we have.
V supply – e 1 = i ⋅ ( R – ∆R )
V supply
e 1 = ---------------------------( 2 – ∆R ⁄ R )
which yields
e1 – 0 = i ⋅ ( R )
The same kind of analysis gives, for the current flow through the two resistors on the right:
V supply
e 2 = ---------------------------( 2 + ∆R ⁄ R )
The output of the bridge, the voltage difference e1 - e2
.e 1
is then
( 2 + ∆R ⁄ R ) – ( 2 – ∆R ⁄ R )
2 ⋅ ( ∆R ⁄ R )
– e 2 = V supply ⋅ ---------------------------------------------------------------- = ------------------------------------V supply
2
( 2 + ∆R ⁄ R ) ⋅ ( 2 – ∆R ⁄ R )
[ 4 – ( ∆R ⁄ R ) ]
But the ratio (∆R/R)2 is going to be much less than 1.0, so we neglect this ratio with respect to 4.0 in
the denominator and obtain the same relationship as in our tension test:
V supply
e 1 – e 2 = ---------------( ∆R ⁄ R )
2
The change in resistance is proportional to the strain, that is
where the gage
∆R ⁄ R = F gage ε
factor is 2.07 So knowing the output voltage (difference) e1 - e2 we can use these last two equations
to determine the strain, ε. The voltage (difference) e1 - e2 is obtained from the output of the opamp, reduced by the Gain factor, i.e., e1 - e2 = (Op-amp output)/(Op-amp Gain)
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Appendix C. - Notes on the use of a spreadsheet.
You are encouraged to make use of a spreadsheet in processing your data and presentation of your
results. However, if you choose to do so, you must make sure you remain in control of your presen­
tation and make clear all steps in your data reduction and analysis. Be careful to:
• Include only the number of significant figures you can justify;
• Label all columns;
• Make explicit on the spreadsheet itself, or in text that accompanies the spreadsheet, the relationships that relate one column to another.
• On a graph generated by the spreadsheet, label all axes, showing units as well as
values at points along the axes that will enable a reader to easily read the plot;
• Do not rely upon color to differentiate among different plots on the same graph. (Unless you want to go to the expense of printing out the graph in color);
• In most cases, the spreadsheet reduction and analysis of data belongs in an appen­
dix. The graph of results, on the other hand, goes in the results section of the main body of a report1. 1. In the structure labs 1 and 2, you are not required to write a full report but only “fill in the
blanks” of the handout. If you use a spreadsheet you may insert the spreadsheet at the appro­
priate place in the handout.Your design task will be documented differently - to be explained in
class.
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